Let ∠RTS=∠RST = a (say)
∠QUA=∠QSU= b(say)
then we know , at point S, a+40+b=180. so, a+b=140 we'll use this later.
consider trianglePQR, ∠P+∠Q+∠R=180
i.e.P+(180-2b)+(180-2a)=180
P+180+180-2(a+b)=180 ⇒P+180-2(a+b)=0 ⇒P=2(140)- 180=280-180=100
hence,answer is E
The first one is 135 per hour if you barely went 90 miles in a hour in a half, and the second one it is 1/3+1/3+1/3 you add 2 more thirds so it means you add 2 forths 1/4/+1/4+1/4=3/4 so you paint 3/4's of the wall
Answer:
∠EBF = 51°
∠DBE = 17°
∠ABF = 141°
∠EBA = 90°
∠DBC = 107°
∠DBF = 68°
Step-by-step explanation:
Hope this helps
Answer:
29.49% probability that a production time is between 9.7 and 12 minutes
Step-by-step explanation:
An uniform probability is a case of probability in which each outcome is equally as likely.
For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.
The probability that we find a value X between c and d, in which d is greater than c, is given by the following formula.
Production times are evenly distributed between 8 and 15.8 minutes and production times are never outside of this interval.
This means that 
What is the probability that a production time is between 9.7 and 12 minutes?
.
So
29.49% probability that a production time is between 9.7 and 12 minutes
Let's call the stamps A, B, and C. They can each be used only once. I assume all 3 must be used in each possible arrangement.
There are two ways to solve this. We can list each possible arrangement of stamps, or we can plug in the numbers to a formula.
Let's find all possible arrangements first. We can easily start spouting out possible arrangements of the 3 stamps, but to make sure we find them all, let's go in alphabetical order. First, let's look at the arrangements that start with A:
ABC
ACB
There are no other ways to arrange 3 stamps with the first stamp being A. Let's look at the ways to arrange them starting with B:
BAC
BCA
Try finding the arrangements that start with C:
C_ _
C_ _
Or we can try a little formula; y×(y-1)×(y-2)×(y-3)...until the (y-x) = 1 where y=the number of items.
In this case there are 3 stamps, so y=3, and the formula looks like this: 3×(3-1)×(3-2).
Confused? Let me explain why it works.
There are 3 possibilities for the first stamp: A, B, or C.
There are 2 possibilities for the second space: The two stamps that are not in the first space.
There is 1 possibility for the third space: the stamp not used in the first or second space.
So the number of possibilities, in this case, is 3×2×1.
We can see that the number of ways that 3 stamps can be attached is the same regardless of method used.
